HDU5480：Conturbatio（前缀和）

最新推荐文章于 2018-02-10 14:12:08 发布

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最新推荐文章于 2018-02-10 14:12:08 发布

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原文链接：http://www.cnblogs.com/junior19/p/6730098.html

本文介绍了一种算法解决棋盘上的攻击问题。该问题涉及多个车（rook）放置在一个棋盘上，每个车可以攻击其所在的行和列。文章通过输入测试案例的数量、棋盘的大小、车的数量及位置等信息，利用累积求和的方法高效判断询问的矩形区域内是否每个格子都被至少一个车攻击。

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Conturbatio

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1035 Accepted Submission(s): 472

Problem Description

There are many rook on a chessboard, a rook can attack the row and column it belongs, including its own place.

There are also many queries, each query gives a rectangle on the chess board, and asks whether every grid in the rectangle will be attacked by any rook?

Input

The first line of the input is a integer

T, meaning that there are

T test cases.

Every test cases begin with four integers

n,m,K,Q.

K is the number of Rook,

Q is the number of queries.

Then

K lines follow, each contain two integers

x,y describing the coordinate of Rook.

Then

Q lines follow, each contain four integers

x1,y1,x2,y2 describing the left-down and right-up coordinates of query.

1≤n,m,K,Q≤100,000.

1≤x≤n,1≤y≤m.

1≤x1≤x2≤n,1≤y1≤y2≤m.

Output

For every query output "Yes" or "No" as mentioned above.

Sample Input

   2 2 2 1 2 1 1 1 1 1 2 2 1 2 2 2 2 2 1 1 1 1 2 2 1 2 2
  

Sample Output

   Yes No Yes 
   
     Hint
    
Huge input, scanf recommended.

Source

BestCoder Round #57 (div.2)

详情请参考：http://blog.youkuaiyun.com/loiecarsers/article/details/48862755

# include <stdio.h>
# include <string.h>
# define MAXN 100000
int main()
{
    int cx, cy, x1, x2, y1, y2, i, t, n, m, k, q, x[MAXN+1], y[MAXN+1];
    scanf("%d",&t);
    while(t--)
    {
        memset(x, 0, sizeof(x));
        memset(y, 0, sizeof(y));
        scanf("%d%d%d%d",&n,&m,&k,&q);
        while(k--)
        {
            scanf("%d%d",&cx,&cy);
            x[cx] = 1;
            y[cy] = 1;
        }
        for(i=2; i<=n; ++i)
            x[i] += x[i-1];
        for(i=2; i<=m; ++i)
            y[i] += y[i-1];
        while(q--)
        {
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            if(x2-x1+1 == x[x2]-x[x1-1] || y2-y1+1 == y[y2]-y[y1-1])
                puts("Yes");
            else
                puts("No");
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/junior19/p/6730098.html